#set document(title: "13.4 Wilcoxon Signed-Rank Test", author: "Rachel Webb")
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== 13.4#h(0.6em)Wilcoxon Signed-Rank Test


    The Wilcoxon Signed-Rank Sum test is the non-parametric alternative to the dependent t-test. The Wilcoxon Signed-Rank Sum test compares the medians of two dependent distributions. The Signed-Rank Sum test, developed by Frank Wilcoxon, finds the difference between paired data values and ranks the absolute value of the differences. Then we sum the ranks for all the negative and positive differences separately. The absolute value of the smaller of these summed ranks is called #math.equation(block: false, alt: "w sub s")[$w_(s)$]. If there were any differences of zero you would not count them in your sample size.


    #figure(figph[Black-and-white portrait photograph of Frank Wilcoxon.], alt: "Black-and-white portrait photograph of Frank Wilcoxon.", caption: [#link("https://en.Wikipedia.org/wiki/Frank_Wilcoxon")[Frank Wilcoxon]])


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Small Sample Size Case: #math.equation(block: false, alt: "n less than 30")[$n < 30$]]


      When the sample size is less than 30, the test statistic is #math.equation(block: false, alt: "w sub s")[$w_(s)$], the absolute value of the smaller of the sum of ranks. Figure 13-5 provides critical values for the Wilcoxon Signed-Rank test. If the test statistic #math.equation(block: false, alt: "w sub s")[$w_(s)$] is greater than the critical value from the table, we fail to reject #math.equation(block: false, alt: "H sub 0")[$H_(0)$]. If the test statistic #math.equation(block: false, alt: "w sub s")[$w_(s)$] is less than or equal to the critical value from the table, we reject #math.equation(block: false, alt: "H sub 0")[$H_(0)$].
]


    #figure(table(
  columns: 8, align: left, inset: 6pt,
  [], [1-Tailed #math.equation(block: false, alt: "α")[$α$]], [], [2-Tailed #math.equation(block: false, alt: "α")[$α$]], [], [], [], [],
  [#math.equation(block: false, alt: "n")[$n$]], [0.01], [0.05], [0.10], [], [0.01], [0.05], [0.10],
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  [6], [-], [2], [3], [], [-], [0], [2],
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  [17], [27], [41], [48], [], [23], [34], [41],
  [18], [32], [47], [55], [], [27], [40], [47],
  [19], [37], [53], [62], [], [32], [46], [53],
  [20], [43], [60], [69], [], [37], [52], [60],
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  [22], [55], [75], [86], [], [48], [65], [75],
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  [28], [101], [130], [145], [], [91], [116], [130],
  [29], [110], [140], [157], [], [100], [126], [140],
))


    #examplebox("Example 1")[][
In an effort to increase production of an automobile part, the factory manager decides to play music in the manufacturing area. Eight workers are selected, and the number of items each produced for a specific day is recorded. After one week of music, the same workers are monitored again. The data are given in the table. At #math.equation(block: false, alt: "α equals 0.05")[$α = 0.05$], can the manager conclude that listening to music has increased production? Use the Wilcoxon Signed-Rank Test since there is no mention of the population being normally distributed.


        Worker 1 2 3 4 5 6 7 8 9


        Before 6 8 10 9 5 12 9 5 7


        After 10 12 9 12 8 13 8 5 10


      
      #solutionbox[
The correct hypotheses are:


        #math.equation(block: false, alt: "H sub 0")[$H_(0)$]: Music in the manufacturing area does not increase production. #linebreak() 
#math.equation(block: false, alt: "H sub 1")[$H_(1)$]: Music in the manufacturing area increases production.


        This is a left-tailed test.


        In order to compute the t-test statistic, first compute the differences between each of the matched pairs.


        #figure(table(
  columns: 10, align: left, inset: 6pt,
  table.header([Before #math.equation(block: false, alt: "open parenthesis x sub 1 close parenthesis")[$( x_(1) )$]], [6], [8], [10], [9], [5], [12], [9], [5], [7]),
  [After #math.equation(block: false, alt: "open parenthesis x sub 2 close parenthesis")[$( x_(2) )$]], [10], [12], [9], [12], [8], [13], [8], [5], [10],
  [#math.equation(block: false, alt: "D equals x sub 1 minus x sub 2")[$D = x_(1) − x_(2)$]], [–4], [–4], [1], [–3], [–3], [–1], [1], [0], [–3],
))


        Take the absolute value of each difference.


        #figure(table(
  columns: 10, align: left, inset: 6pt,
  table.header([Before #math.equation(block: false, alt: "open parenthesis x sub 1 close parenthesis")[$( x_(1) )$]], [6], [8], [10], [9], [5], [12], [9], [5], [7]),
  [After #math.equation(block: false, alt: "open parenthesis x sub 2 close parenthesis")[$( x_(2) )$]], [10], [12], [9], [12], [8], [13], [8], [5], [10],
  [#math.equation(block: false, alt: "D equals x sub 1 minus x sub 2")[$D = x_(1) − x_(2)$]], [–4], [–4], [1], [–3], [–3], [–1], [1], [0], [–3],
  [#math.equation(block: false, alt: "vertical bar D vertical bar")[$| D |$]], [4], [4], [1], [3], [3], [1], [1], [0], [3],
))


        Rank the data and drop any ties. At this point, if any of the differences are zero, that pair is no longer used and is not ranked.


        #figure(table(
  columns: 10, align: left, inset: 6pt,
  table.header([Before #math.equation(block: false, alt: "open parenthesis x sub 1 close parenthesis")[$( x_(1) )$]], [6], [8], [10], [9], [5], [12], [9], [5], [7]),
  [After #math.equation(block: false, alt: "open parenthesis x sub 2 close parenthesis")[$( x_(2) )$]], [10], [12], [9], [12], [8], [13], [8], [5], [10],
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  [#math.equation(block: false, alt: "vertical bar D vertical bar")[$| D |$]], [4], [4], [1], [3], [3], [1], [1], [0], [3],
  [Rank], [7.5], [7.5], [2], [5], [5], [2], [2], [drop], [5],
))


        The sample size #math.equation(block: false, alt: "n")[$n$] is the number of differences that are not zero. So, in this case, #math.equation(block: false, alt: "n equals 8")[$n = 8$]. Next, take the sign of the difference and attach this plus or minus sign to each rank.


        #figure(table(
  columns: 9, align: left, inset: 6pt,
  table.header([Before #math.equation(block: false, alt: "open parenthesis x sub 1 close parenthesis")[$( x_(1) )$]], [6], [8], [10], [9], [5], [12], [9], [7]),
  [After #math.equation(block: false, alt: "open parenthesis x sub 2 close parenthesis")[$( x_(2) )$]], [10], [12], [9], [12], [8], [13], [8], [10],
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  [#math.equation(block: false, alt: "vertical bar D vertical bar")[$| D |$]], [4], [4], [1], [3], [3], [1], [1], [3],
  [Rank], [7.5], [7.5], [2], [5], [5], [2], [2], [5],
  [Signed Rank], [–7.5], [–7.5], [+2], [–5], [–5], [-2], [+2], [–5],
))


        Find the sum of the positive and negative ranks:


        Positive ranks: #math.equation(block: false, alt: "2 plus 2 equals 4")[$2 + 2 = 4$]


        Negative ranks: #math.equation(block: false, alt: "open parenthesis minus 7.5 close parenthesis plus open parenthesis minus 7.5 close parenthesis plus open parenthesis minus 5 close parenthesis plus open parenthesis minus 5 close parenthesis plus open parenthesis minus 2 close parenthesis plus open parenthesis minus 5 close parenthesis equals minus 32")[$( − 7.5 ) + ( − 7.5 ) + ( − 5 ) + ( − 5 ) + ( − 2 ) + ( − 5 ) = − 32$].


        Take the smaller of the absolute value of the sums of the ranks: #math.equation(block: false, alt: "vertical bar 4 vertical bar equals 4 , vertical bar minus 32 vertical bar equals 32")[$| 4 | = 4 , | − 32 | = 32$], so 4 is smaller.


        This is our test statistic called #math.equation(block: false, alt: "w sub s equals 4")[$w_(s) = 4$].


        Next, use the table in Figure 13-5 to get the critical value. The table provides critical values for two-tailed tests.


        This is a one-tailed test, with #math.equation(block: false, alt: "α equals 0.05")[$α = 0.05$] and #math.equation(block: false, alt: "n equals 8")[$n = 8$]. See Figure 13-6 that shows which row and column from Figure 13- 5 to use to find the critical value.


        #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([], [], [1-Tailed #math.equation(block: false, alt: "α")[$α$]], []),
  [#math.equation(block: false, alt: "n")[$n$]], [0.01], [0.05], [0.10],
  [5], [-], [0], [2],
  [6], [-], [2], [3],
  [7], [0], [3], [5],
  [8], [1], [5], [8],
))


        The critical value = 5. The test statistic #math.equation(block: false, alt: "w sub s equals 4")[$w_(s) = 4$] is less than the critical value of 5.


        The decision rule for the critical values in Figure 13-5 is to reject the null if the test statistic is less than or equal to the critical value, and do not reject the null hypothesis if the test statistic is larger than the critical value.


        Since #math.equation(block: false, alt: "w sub s less than Critical Value equals 5")[$w_(s) < "Critical Value" = 5$], the decision is to reject #math.equation(block: false, alt: "H sub 0")[$H_(0)$]. There is enough evidence to support the claim that listening to music has increased production.


        When the sample size is 30 or more,a paired t-test may be used in most situations. However, if your population is heavily skewed or you are using interval data, then use the large sample size normal approximation Wilcoxon Signed-Rank test.
]
]


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Large Sample Size Case: #math.equation(block: false, alt: "n greater than or equal to 30")[$n ≥ 30$]]


      We can use the normal approximation for sample sizes of 30 or more. The formula for the test statistic is: #math.equation(block: true, alt: "z equals the fraction open parenthesis w sub s minus open parenthesis the fraction n open parenthesis n plus 1 close parenthesis over 4 close parenthesis close parenthesis over the square root of open parenthesis the fraction n open parenthesis n plus 1 close parenthesis open parenthesis 2 n plus 1 close parenthesis over 24 close parenthesis")[$z = frac(( w_(s) − ( frac(n ( n + 1 ), 4) ) ), sqrt(( frac(n ( n + 1 ) ( 2 n + 1 ), 24) )))$]


      where #math.equation(block: false, alt: "n")[$n$] is the reduced samples size excluding any differences of zero, and #math.equation(block: false, alt: "w sub s")[$w_(s)$] is the smaller in absolute value of the signed ranks for a two-tailed test, the sum of the positive ranks for a left-tailed test, or the sum of the negative ranks for a right-tailed test. The sample size #math.equation(block: false, alt: "n")[$n$] is the reduced sample size not including any differences of zero.
]


    #examplebox("Example 2")[][
A pharmaceutical company is testing to see if there is a significant difference in the pain relief for two new pain medications. They randomly assign the two different pain medications for 34 patients with chronic pain and record the pain rating for each patient one hour after each dose. The pain ratings are on a sliding scale from 1 to 10. The results are listed below. Use the Wilcoxon Signed-Rank test to see if there is a significant difference at #math.equation(block: false, alt: "α equals 0.05")[$α = 0.05$].


        #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Finish the large-sample Wilcoxon on the normal curve]


          The ranking is by hand, but everything after w\_s = 128.5 and n = 31 is a standard-normal read. The first link opens inverse mode at area 0.025 and returns the book's critical value of -1.96 (so +/-1.96). The second shades the tail below the test statistic z = -2.3418; it lands outside the critical value, and doubling 0.0096 gives the matching two-tailed p-value 0.0192 that the book leaves unprinted.


          - invNorm(0.025) = -1.96
- P(Z \< -2.3418) = 0.0096 (doubled = 0.0192)
]


        
=== Patient Drug 1 Drug 2 Patient Drug 1 Drug 2 Patient Drug 1 Drug 2 1 2.4 2.5   13 4 6.1   25 4 5.1 2 4.7 3.3   14 2.2 2.9   26 5.5 4.4 3 1.2 5.3   15 2.7 4.3   27 3.6 3.6 4 5.9 5.6   16 2.9 3.3   28 3.8 3.5 5 4.5 5   17 5 5   29 5.4 4.8 6 4 5.3   18 3.1 5.1   30 2.4 3.2 7 2.5 4.6   19 3.3 3.3   31 4.1 2.6 8 3 2.5   20 3 5.9   32 4.5 5.7 9 5 3.4   21 5.4 3.2   33 4 5.8 10 5.8 5.4   22 4.2 5.9   34 6 5 11 1.9 5.1   23 3.6 5.9         12 3.2 4.3   24 2.2 5.6        Solution


      
      #solutionbox[
The correct hypotheses are:


        #math.equation(block: false, alt: "H sub 0")[$H_(0)$]: There is no difference in the pain scale rating for the two pain medications. #linebreak() 
#math.equation(block: false, alt: "H sub 1")[$H_(1)$]: There is a difference in the pain scale rating for the two pain medications.


        Compute the differences between each of the matched pairs. Rank the absolute value of the differences. Make sure to average the ranks repeated differences and do not rank any differences of zero. After the differences are ranked, attach this sign of the difference to each rank.


        #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([Patient], [Drug 1], [Drug 2], [Difference], [#math.equation(block: false, alt: "vertical bar D vertical bar")[$| D |$]], [Rank], [Signed Rank]),
  [1], [2.4], [2.5], [–0.1], [0.1], [1], [–1],
  [2], [4.7], [3.3], [1.4], [1.4], [17], [17],
  [3], [1.2], [5.3], [–4.1], [4.1], [31], [–31],
  [4], [5.9], [5.6], [0.3], [0.3], [2.5], [2.5],
  [5], [4.5], [5], [–0.5], [0.5], [6.5], [–6.5],
  [6], [4], [5.3], [–1.3], [1.3], [16], [–16],
  [7], [2.5], [4.6], [–2.1], [2.1], [24.5], [–24.5],
  [8], [3], [2.5], [0.5], [0.5], [6.5], [6.5],
  [9], [5], [3.4], [1.6], [1.6], [19.5], [19.5],
  [10], [5.8], [5.4], [0.4], [0.4], [4.5], [4.5],
  [11], [1.9], [5.1], [–3.2], [3.2], [29], [–29],
  [12], [3.2], [4.3], [–1.1], [1.1], [13], [–13],
  [13], [4], [6.1], [–2.1], [2.1], [24.5], [–24.5],
  [14], [2.2], [2.9], [–0.7], [0.7], [9], [–9],
  [15], [2.7], [4.3], [–1.6], [1.6], [19.5], [–19.5],
  [16], [2.9], [3.3], [–0.4], [0.4], [4.5], [–4.5],
  [17], [5], [5], [0], [], [], [],
  [18], [3.1], [5.1], [–2], [2], [23], [–23],
  [19], [3.3], [3.3], [0], [], [], [],
  [20], [3], [5.9], [–2.9], [2.9], [28], [–28],
  [21], [5.4], [3.2], [2.2], [2.2], [26], [26],
  [22], [4.2], [5.9], [–1.7], [1.7], [21], [–21],
  [23], [3.6], [5.9], [–2.3], [2.3], [27], [–27],
  [24], [2.2], [5.6], [–3.4], [3.4], [30], [–30],
  [25], [4], [5.1], [–1.1], [1.1], [13], [– 13],
  [26], [5.5], [4.4], [1.1], [1.1], [13], [13],
  [27], [3.6], [3.6], [0], [], [], [],
  [28], [3.8], [3.5], [0.3], [0.3], [2.5], [2.5],
  [29], [5.4], [4.8], [0.6], [0.6], [8], [8],
  [30], [2.4], [3.2], [–0.8], [0.8], [10], [–10],
  [31], [4.1], [2.6], [1.5], [1.5], [18], [18],
  [32], [4.5], [5.7], [–1.2], [1.2], [15], [–15],
  [33], [4], [5.8], [–1.8], [1.8], [22], [–22],
  [34], [6], [5], [1], [1], [11], [11],
))


        Find the sum of the positive and negative ranks:


        Positive ranks: #math.equation(block: false, alt: "17 plus 2.5 plus 6.5 plus 19.5 plus 4.5 plus 26 plus 13 plus 2.5 plus 8 plus 18 plus 11 equals 128.5")[$17 + 2.5 + 6.5 + 19.5 + 4.5 + 26 + 13 + 2.5 + 8 + 18 + 11 = 128.5$]


        Negative ranks: #math.equation(block: false, alt: "open parenthesis minus 1 close parenthesis plus open parenthesis minus 31 close parenthesis plus open parenthesis minus 6.5 close parenthesis plus open parenthesis minus 16 close parenthesis plus open parenthesis minus 24.5 close parenthesis plus open parenthesis minus 29 close parenthesis plus open parenthesis minus 13 close parenthesis plus open parenthesis minus 24.5 close parenthesis plus open parenthesis minus 9 close parenthesis plus open parenthesis minus 19.5 close parenthesis plus open parenthesis minus 4.5 close parenthesis plus open parenthesis minus 23 close parenthesis plus open parenthesis minus 28 close parenthesis plus open parenthesis minus 21 close parenthesis plus open parenthesis minus 27 close parenthesis plus open parenthesis minus 30 close parenthesis plus open paren")[$( − 1 ) + ( − 31 ) + ( − 6.5 ) + ( − 16 ) + ( − 24.5 ) + ( − 29 ) + ( − 13 ) + ( − 24.5 ) + ( − 9 ) + ( − 19.5 ) + ( − 4.5 ) + ( − 23 ) + ( − 28 ) + ( − 21 ) + ( − 27 ) + ( − 30 ) + ( − 13 ) + ( − 10 ) + ( − 15 ) + ( − 22 ) = − 367.5$].


        Take the smaller of the absolute value of the sums of the ranks: #math.equation(block: false, alt: "vertical bar 128.5 vertical bar equals 128.5 , vertical bar minus 367.5 vertical bar equals 367.5")[$| 128.5 | = 128.5 , | − 367.5 | = 367.5$], so 128.5 is smaller.


        The smaller of the absolute value of the sum of the ranks is #math.equation(block: false, alt: "w sub s equals 128.5")[$w_(s) = 128.5$].


        Throw out the three differences of zero. The sample size is #math.equation(block: false, alt: "n equals 31")[$n = 31$].


        The test statistic is: #math.equation(block: false, alt: "z equals the fraction open parenthesis w sub s minus open parenthesis the fraction n open parenthesis n plus 1 close parenthesis over 4 close parenthesis close parenthesis over the square root of open parenthesis the fraction n open parenthesis n plus 1 close parenthesis open parenthesis 2 n plus 1 close parenthesis over 24 close parenthesis equals the fraction open parenthesis 128.5 minus open parenthesis the fraction 31 times 32 over 4 close parenthesis close parenthesis over the square root of open parenthesis the fraction 31 times 32 times 63 over 24 close parenthesis equals the fraction open parenthesis 128.5 minus 248 close parenthesis over the square root of open parenthesis 2604 close parenthesis equals minus 2.341787")[$z = frac(( w_(s) − ( frac(n ( n + 1 ), 4) ) ), sqrt(( frac(n ( n + 1 ) ( 2 n + 1 ), 24) ))) = frac(( 128.5 − ( frac(31 · 32, 4) ) ), sqrt(( frac(31 · 32 · 63, 24) ))) = frac(( 128.5 − 248 ), sqrt(( 2604 ))) = − 2.341787$]


        Either the #math.equation(block: false, alt: "z")[$z$] critical value or p-value method may be used, similar to how we used previous z-tests.


        Compute the #math.equation(block: false, alt: "z sub α / 2")[$z_(α / 2)$] critical values. Draw and label the distribution; see Figure 13-7.


        #figure(figph[z\_alpha/2 critical values of positive and negative 1.96.], alt: "z_alpha/2 critical values of positive and negative 1.96.", caption: [Figure 13-7: #math.equation(block: false, alt: "z sub α / 2")[$z_(α / 2)$] critical values.])


        Use the inverse normal function #math.equation(block: false, alt: "invNorm open parenthesis 0.025 , 0 , 1 close parenthesis")[$"invNorm" ( 0.025 , 0 , 1 )$] to get #math.equation(block: false, alt: "z sub α / 2 equals ± 1.96")[$z_(α / 2) = ± 1.96$]. The test statistic #math.equation(block: false, alt: "z equals minus 2.3418")[$z = − 2.3418$] is in the shaded critical region, so reject #math.equation(block: false, alt: "H sub 0")[$H_(0)$].


        There is enough evidence to support the claim that there is a significant difference in the pain scale rating for the two pain medications.


        These calculations can be done by hand or using the following online calculator: #strong[#link("http://www.socscistatistics.com/tests/signedranks")[http://www.socscistatistics.com/tests/signedranks]].


        #strong[The TI calculators and Excel do not have built-in nonparametric tests.]
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    It is an important and popular fact that things are not always what they seem. For instance, on the planet Earth, man had always assumed that he was more intelligent than dolphins because he had achieved so much – the wheel, New York, wars and so on – whilst all the dolphins had ever done was muck about in the water having a good time. But conversely, the dolphins had always believed that they were far more intelligent than man – for precisely the same reasons.


    (Adams, 2002)